Problem 52 - University Physics with Modern Physics

Question

The position of a training helicopter (weight 2.75 \(\times\) 10\(^5\) N) in a test is given by \(\hat{r}\) = (0.020 m/s\(^3)t^3\hat{i}\) + (2.2 m/s) \(t\hat{j}\) - (0.060 m/s\(^2)t^2\hat{k}\) n. Find the net force on the helicopter at \(t =\) 5.0 s.

Step-by-Step Solution

Answer

The net force on the helicopter at \( t = 5.0 \) s is approximately \( \vec{F} = (16820.18 \hat{i} - 3364.04 \hat{k}) \text{ N} \).

Step 1: Understanding the Problem

We need to find the net force on the helicopter at time \( t = 5.0 \) s. The position of the helicopter is given as a function of time, and we know the helicopter's weight. We will use the relation \( \vec{F} = m \vec{a} \), where \( \vec{a} \) is the acceleration, to find the net force.

Step 2: Find the Velocity by Differentiating Position

The velocity \( \vec{v} \) is the first derivative of position \( \vec{r} \) with respect to time \( t \). Differentiate each component of \( \vec{r} \):- \( \frac{d}{dt}(0.020 \text{ m/s}^3 t^3) = 0.060 \text{ m/s}^3 t^2 \).- \( \frac{d}{dt}(2.2 \text{ m/s} \cdot t) = 2.2 \text{ m/s} \).- \( \frac{d}{dt}(-0.060 \text{ m/s}^2 t^2) = -0.120 \text{ m/s}^2 t \).

Step 3: Find the Acceleration by Differentiating Velocity

The acceleration \( \vec{a} \) is the first derivative of velocity \( \vec{v} \). Differentiate each velocity component:- \( \frac{d}{dt}(0.060 \text{ m/s}^3 t^2) = 0.120 \text{ m/s}^3 t \).- \( \frac{d}{dt}(2.2 \text{ m/s}) = 0 \).- \( \frac{d}{dt}(-0.120 \text{ m/s}^2 t) = -0.120 \text{ m/s}^2 \).

Step 4: Evaluate Acceleration at t = 5.0 s

Substitute \( t = 5.0 \) s into the acceleration components we found:- \( a_x = 0.120 \text{ m/s}^3 \times 5.0 \text{ s} = 0.600 \text{ m/s}^2 \).- \( a_y = 0 \text{ m/s}^2 \).- \( a_z = -0.120 \text{ m/s}^2 \).

Step 5: Calculate the Helicopter's Mass

Given the weight \( W = 2.75 \times 10^5 \) N and using \( W = mg \) where \( g = 9.81 \text{ m/s}^2 \), we calculate mass \( m \):\[ m = \frac{W}{g} = \frac{2.75 \times 10^5 \text{ N}}{9.81 \text{ m/s}^2} \approx 28033.64 \text{ kg} \].

Step 6: Calculate the Net Force

Use \( \vec{F} = m \vec{a} \) to find the net force on the helicopter:- \( F_x = m \cdot a_x = 28033.64 \text{ kg} \times 0.600 \text{ m/s}^2 = 16820.18 \text{ N} \).- \( F_y = m \cdot a_y = 28033.64 \text{ kg} \times 0 = 0 \text{ N} \).- \( F_z = m \cdot a_z = 28033.64 \text{ kg} \times (-0.120 \text{ m/s}^2) = -3364.04 \text{ N} \).

Step 7: Express the Net Force as a Vector

The net force \( \vec{F} \) is given by the vector sum of its components: \[ \vec{F} = (16820.18 \hat{i} + 0 \hat{j} - 3364.04 \hat{k}) \text{ N} \].

Key Concepts

Acceleration from Position Function

Understanding acceleration through position functions allows us to solve many physics problems, especially when motion is involved. To find acceleration from a position function like the one given for the helicopter, we must differentiate the position vector function twice. First, the position function \( \hat{r} = (0.020 \text{ m/s}^3 t^3 \hat{i}) + (2.2 \text{ m/s} \cdot t \hat{j}) - (0.060 \text{ m/s}^2 t^2 \hat{k}) \) is differentiated with respect to time to find velocity. Remember, differentiation is the act of finding the rate of change. This gives us:

• Velocity of the x-component: \( v_x = 0.060 \text{ m/s}^3 t^2 \)
• Velocity of the y-component: \( v_y = 2.2 \text{ m/s} \)
• Velocity of the z-component: \( v_z = -0.120 \text{ m/s}^2 t \)

Next, we differentiate the velocity components to get acceleration:

• Acceleration of x-component: \( a_x = 0.120 \text{ m/s}^3 t \)
• Acceleration of y-component: \( a_y = 0 \)
• Acceleration of z-component: \( a_z = -0.120 \text{ m/s}^2 \)

This step reveals how different aspects of movement contribute to the total acceleration of an object, like our helicopter.

Differentiation in Physics

Differentiation is like a magic wand in physics that helps us understand changes in motion and other physical properties. It is critical when dealing with functions that depend on time, like our helicopter's position. By taking derivatives, we go from understanding where something is to how fast it's moving, and then to how its speed changes over time. In this problem, we perform differentiation in two key steps:

• First, find velocity: This is the rate of change of position over time. By differentiating the position function, you can see how fast the position changes.
• Second, find acceleration: This is the rate of change of velocity over time. Differentiating the velocity function shows us how rapidly the speed or direction changes.

Differentiation helps break down complex motion into manageable parts, allowing us to analyze and understand even the most dynamic systems.

Helicopter Dynamics

Helicopter dynamics can seem complex, but they are manageable when you understand the basics of motion and forces. For a training helicopter in tests like this, the key is understanding how forces act and cause movement. The helicopter's weight provides a constant downward force due to gravity, calculated using its mass and the gravitational constant, \( g = 9.81 \text{ m/s}^2 \). In our exercise, the helicopter’s mass was found using its weight. This mass is then essential when finding the net force from the calculated acceleration. Helicopters can have complex motion involving translation and rotation, but basic dynamics start with:

• Understanding its position and speed changes
• Calculating forces acting on it
• Using these forces to analyze how the helicopter will move

When we find the acceleration and apply Newton's Second Law, \( \vec{F} = m \vec{a} \), we're able to see how these dynamics play out in a quantifiable manner.

Vector Components in Forces

Working with forces in multiple dimensions requires understanding vectors, as forces have both magnitude and direction. Consider the helicopter we are analyzing. The force vector is broken into components along the x, y, and z axes, which correspond to different directional forces acting on the helicopter: horizontal, vertical, and depth.

• Net force in the x-direction, calculated from \( F_x = m \cdot a_x \)
• No force in the y-direction, as \( F_y = 0 \).
• Net force in the z-direction, calculated from \( F_z = m \cdot a_z \)

The net force vector allows us to sum these up and visualize the total force as:\[ \vec{F} = (16820.18 \hat{i} + 0 \hat{j} - 3364.04 \hat{k}) \text{ N} \]Vectors break down complex interactions into understandable parts, helping us clarity the motion and interaction of forces on our helicopter.